1 chemical equilibrium chapter 15 copyright © the mcgraw-hill companies, inc. permission required...
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Chemical EquilibriumChapter 15
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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Equilibrium is a state in which there are no observable changes as time goes by.
Chemical equilibrium is achieved when:
• the rates of the forward and reverse reactions are equal and
• the concentrations of the reactants and products remain constant
Physical equilibrium
H2O (l)
Chemical equilibrium
N2O4 (g)
H2O (g)
2NO2 (g)
NO2
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N2O4 (g) 2NO2 (g)
Start with NO2 Start with N2O4 Start with NO2 & N2O4
equilibrium
equilibrium equilibrium
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constant
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N2O4 (g) 2NO2 (g)
= 4.63 x 10-3K = [NO2]2
[N2O4]
aA + bB cC + dD
K = [C]c[D]d
[A]a[B]bLaw of Mass Action
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K >> 1
K << 1
Lie to the right Favor products
Lie to the left Favor reactants
Equilibrium Will
K = [C]c[D]d
[A]a[B]baA + bB cC + dD
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Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
N2O4 (g) 2NO2 (g)
Kc = [NO2]2
[N2O4]Kp =
NO2P 2
N2O4P
aA (g) + bB (g) cC (g) + dD (g)
Kp = Kc(RT)n
n = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)
In most cases
Kc Kp
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Homogeneous Equilibrium
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
Kc =′[CH3COO-][H3O+][CH3COOH][H2O]
[H2O] = constant
Kc = [CH3COO-][H3O+]
[CH3COOH]= Kc [H2O]′
General practice not to include units for the equilibrium constant.
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The equilibrium constant Kp for the reaction
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm?2
2NO2 (g) 2NO (g) + O2 (g)
Kp = 2PNO PO
2
PNO2
2
PO2 = Kp
PNO2
2
PNO2
PO2 = 158 x (0.400)2/(0.270)2 = 347 atm
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Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.
CaCO3 (s) CaO (s) + CO2 (g)
[CaCO3] = constant[CaO] = constant
Kc = [CO2] = Kp = PCO2
The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.
[CaO][CO2][CaCO3]
Kc =′
[CaCO3][CaO]
Kc x′
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PCO 2= Kp
CaCO3 (s) CaO (s) + CO2 (g)
PCO 2 does not depend on the amount of CaCO3 or CaO
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Consider the following equilibrium at 295 K:
The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction?
NH4HS (s) NH3 (g) + H2S (g)
Kp = PNH3 H2SP = 0.265 x 0.265 = 0.0702
Kp = Kc(RT)n
Kc = Kp(RT)-n
n = 2 – 0 = 2 T = 295 K
Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4
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N2O4 (g) 2NO2 (g)
= 4.63 x 10-3K = [NO2]2
[N2O4]
2NO2 (g) N2O4 (g)
K = [N2O4]
[NO2]2′ =
1K
= 216
When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.
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Writing Equilibrium Constant Expressions
1. The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.
2. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.
3. The equilibrium constant is a dimensionless quantity.
4. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.
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The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.
IF
• Qc > Kc system proceeds from right to left to reach equilibrium
• Qc = Kc the system is at equilibrium
• Qc < Kc system proceeds from left to right to reach equilibrium
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Calculating Equilibrium Concentrations
1. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.
2. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.
3. Having solved for x, calculate the equilibrium concentrations of all species.
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At 1280oC the equilibrium constant (Kc) for the reaction
Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.
Br2 (g) 2Br (g)
Br2 (g) 2Br (g)
Let x be the change in concentration of Br2
Initial (M)
Change (M)
Equilibrium (M)
0.063 0.012
-x +2x
0.063 - x 0.012 + 2x
[Br]2
[Br2]Kc = Kc =
(0.012 + 2x)2
0.063 - x= 1.1 x 10-3 Solve for x
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Kc = (0.012 + 2x)2
0.063 - x= 1.1 x 10-3
4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x
4x2 + 0.0491x + 0.0000747 = 0
ax2 + bx + c =0-b ± b2 – 4ac
2ax =
Br2 (g) 2Br (g)
Initial (M)
Change (M)
Equilibrium (M)
0.063 0.012
-x +2x
0.063 - x 0.012 + 2x
x = -0.00178x = -0.0105
At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M
At equilibrium, [Br2] = 0.062 – x = 0.0648 M
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If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.
Le Châtelier’s Principle
• Changes in Concentration
N2 (g) + 3H2 (g) 2NH3 (g)
AddNH3
Equilibrium shifts left to offset stress
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Le Châtelier’s Principle
• Changes in Concentration continued
Change Shifts the Equilibrium
Increase concentration of product(s) left
Decrease concentration of product(s) right
Decrease concentration of reactant(s)
Increase concentration of reactant(s) right
left
aA + bB cC + dD
AddAddRemove Remove
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Le Châtelier’s Principle
• Changes in Volume and Pressure
A (g) + B (g) C (g)
Change Shifts the Equilibrium
Increase pressure Side with fewest moles of gas
Decrease pressure Side with most moles of gas
Decrease volume
Increase volume Side with most moles of gas
Side with fewest moles of gas
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Le Châtelier’s Principle
• Changes in Temperature
Change Exothermic Rx
Increase temperature K decreases
Decrease temperature K increases
Endothermic Rx
K increases
K decreases
colder hotter
N2O4 (g) 2NO2 (g)
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Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.
• Adding a Catalyst• does not change K• does not shift the position of an equilibrium system• system will reach equilibrium sooner
Le Châtelier’s Principle
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Le Châtelier’s Principle - Summary
Change Shift EquilibriumChange Equilibrium
Constant
Concentration yes no
Pressure yes* no
Volume yes* no
Temperature yes yes
Catalyst no no
*Dependent on relative moles of gaseous reactants and products
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